Question

Question

An industrial system consisting of three (3) sub units with reliability data listed below in terms of exponential failure rates and MTTR times for a mission time of one week. The system operates 9 hours per day with one day per week for down time. The system data are:

Λ 1 = 0.00020 failure per hour

Λ 2 = 0.00012 failure per hour,

Λ 3 = 0.00040 failure per hour,

MTTR₁ = 1.00 hour,

MTTR₂ = 1.25 hour,

MTTR₃ = 1.625 hour,

a) Calculate the availability of the system when connected in series.

b) Calculate the availability of the system when connected in parallel.

Answer 1

MTBF1 = 1 / Λ1 = 1/0.0002 = 5000 hours

MTBF2 = 1 / Λ2 = 1/0.00012 = 8333.33 hours

MTBF3 = 1 / Λ3 = 1/0.0004 = 2500 hours

MTTR1 = 1.00 hour

MTTR2 = 1.25 hour

MTTR3 = 1.625 hour

Availability 1 = MTBF1 / (MTBF1 + MTTR1) = 5000 / (5000 + 1) = 0.9998

Availability 2 = MTBF2 / (MTBF2 + MTTR2) = 8333.33 / (8333.33 + 1.25) = 0.99985

Availability 3 = MTBF3 / (MTBF3 + MTTR3) = 2500 / (2500 + 1.625) = 0.99935

Availability 1 = 0.9998

Availability 2 = 0.99985

Availability 3 = 0.99935

a.

Availibility of system in series :

Availibility of system = 0.9998*0.99985*0.99935 = 0.99900025748 ≈ 0.999

b.

Availibility of system in parallel :

Availibility of system = 1 - (1-0.9998)*(1-0.99985)*(1-0.99935) = 0.99999999998 ≈ 1

P.S. (please upvote if you find the answer satisfactory)